\(x^4+1+x^2\ge2\sqrt{x^4}+x^2=3x^2\)
\(\Rightarrow M\le\dfrac{x^2}{3x^2}=\dfrac{1}{3}\)
\(\Rightarrow M_{max}=\dfrac{1}{3}\) khi \(x^4=1\Rightarrow x=\pm1\)
\(x^4+1+x^2\ge2\sqrt{x^4}+x^2=3x^2\)
\(\Rightarrow M\le\dfrac{x^2}{3x^2}=\dfrac{1}{3}\)
\(\Rightarrow M_{max}=\dfrac{1}{3}\) khi \(x^4=1\Rightarrow x=\pm1\)
tim gia tri lon nhat cua bieu thuc \(f\left(x\right)=\dfrac{1}{x^4-x^2+1}\)
gia tri nho nhat cua bieu thuc\(\dfrac{^{^{ }}^{ }x^2+1}{x^2-1^{ }}\)
cho a c la cac so thuc khong am thoa man dieu kien a+b+c=3 tim gia tri lon nhat cua bieu thucP=a\b3+1+b\c3+1+c\a3+1
cho x,y>0 va \(x+y\le1.\)
tim GTNN cua bieu thuc \(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}\)
Cho \(xy+\sqrt{\left(1+x\right)^2\left(1+y\right)^2}=\sqrt{2017}\)
Tinh gia tri cua bieu thuc: \(P=x+\sqrt{1+y^2}+y\sqrt{1+x^2}\)
Giup mk!!!
cho phuong trinh:\(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
a/tim dieu kien cua x de phuong trinh co nghia
b/giai phuong trinh
tim gtnn cua bt A
A= \((x-1)^4+(x-3)^4+6(x-1)^2\left(x-3\right)^2\)
A=(\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)):\(\dfrac{\sqrt{x}-1}{2}\)
Voi x≥0 va x≠1
a)Tim x de A=2/7
b)So sanh A^2va 2A
a) RÚT GỌN M
b) TÌM X ĐỂ M=4/5
c) so sánh M và M^2